Automatic Control Systems Solutions Manual Kuo Golnaragh by Benjamin Kuo PDF

By Benjamin Kuo

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4 !   cos t = − sin t sin t  cos t  (2) Inverse Laplace transform: −1 1  s 1  s −1  Φ ( s ) = ( sI − A ) = 1 s  = s 2 + 1  −1 s  −1 φ (t ) =  cos t  − sin t sin t  cos t  (b) A=  −1 0   0 −2  A = 2 1 0   0 4  A = 3  −1 0   0 − 8  A = 4  −1 0   0 16  (1) Infinite series expansion: 2 3 4 t t t   1 − t + − + +L 0   1 2 2 2 ! 3! 4 ! φ ( t) = I + At + A t + L =   2 3 2! 4t 8t   0 1 − 2t + − +L   2! 3! e −t = 0   e  0 −2 t (2) Inverse Laplace transform: Φ ( s ) = ( sI − A ) = −1 s + 1  0  1   s +1 = s + 2   0     1  s + 2  0 −1 0  e− t  −2 t  e  0 φ (t ) =  0 (c) A= 0 1  1 0  A = 2 1 0  0 1  A = 3 0 1  1 0  A = 4 1 0  0 1  (1) Infinite series expansion: 3 5 t t  t2 t4  1+ + +L t+ + L   1 2 2  e− t + et 2!

2! 4 !   cos t = − sin t sin t  cos t  (2) Inverse Laplace transform: −1 1  s 1  s −1  Φ ( s ) = ( sI − A ) = 1 s  = s 2 + 1  −1 s  −1 φ (t ) =  cos t  − sin t sin t  cos t  (b) A=  −1 0   0 −2  A = 2 1 0   0 4  A = 3  −1 0   0 − 8  A = 4  −1 0   0 16  (1) Infinite series expansion: 2 3 4 t t t   1 − t + − + +L 0   1 2 2 2 ! 3! 4 ! φ ( t) = I + At + A t + L =   2 3 2! 4t 8t   0 1 − 2t + − +L   2! 3! e −t = 0   e  0 −2 t (2) Inverse Laplace transform: Φ ( s ) = ( sI − A ) = −1 s + 1  0  1   s +1 = s + 2   0     1  s + 2  0 −1 0  e− t  −2 t  e  0 φ (t ) =  0 (c) A= 0 1  1 0  A = 2 1 0  0 1  A = 3 0 1  1 0  A = 4 1 0  0 1  (1) Infinite series expansion: 3 5 t t  t2 t4  1+ + +L t+ + L   1 2 2  e− t + et 2!

The last column of Thus, the last column of adj (s I − A ) adj (s I − A ) B 50 is 1 adj (s I − A ) is is obtained from the cofactors of s s 2 L s n −1 ' . 0 x 5-18 (a) State variables: x 1 = y, x 2 = dy , dt x 3 = d 2 y 2 dt x& ( t ) = Ax ( t ) + B r ( t ) State equ ations: 0 1 0 A= 0 0 1    −1 −3 −3   0 B =  0   1  (b) State transition matrix: s2 + 3 s + 3 s + 3 1   s −1 0  1  −1  2 Φ ( s ) = ( sI − A ) =  0 s −1  = −1 s + 3s s     ∆ (s ) 2  − s  1 3 s + 3  −3 s − 1 s   1 1 1 1 2 1  + + +   2 3 ( s + 1) 2 ( s + 1)3 ( s + 1 )3   s + 1 ( s + 1) ( s + 1)   −1 1 1 2 s = + −  2 s + 1 ( s + 1) ( s + 1 )3 ( s + 1) 3 ( s + 1) 3     2 −s −3 2 s   +  ( s + 1 )3 ( s + 1) 2 ( s + 1)3 ( s + 1)3  −1 ∆ (s ) = s3 + 3 s2 + 3s + 1 = ( s + 1) (1 + t + t 2 / 2 ) e− t  2 −t φ ( t) =  −t e / 2  ( −t + t 2 / 2 ) e −t (t + t ) e (1 + t − t ) e −t 2 2 2 t e 3  ( t − t / 2 ) e  (1 − 2t + t 2 / 2 ) e −t  2 −t t e /2 −t −t 2 −t (d) Characteristic equation: ∆ ( s ) = s + 3 s + 3 s + 1 = 0 3 − 1, −1, −1 Eigenvalues: 5-19 (a) State variables: 2 x 1 = y, x 2 = dy dt State equations:  dx1 (t )   dt   0 1   x1 ( t)   0   =   +   r ( t)  dx2 ( t)   − 1 − 2   x2 ( t)   1   dt  State transition matrix:  s+ 2  ( s + 1) 2  s −1  −1  Φ ( s ) = ( sI − A ) =  1 s + 2  =  − 1  ( s + 1) 2  −1  ( s + 1)  s  ( s + 1)2  1 2 51  (1 + t ) e−t φ (t ) =   − te −t   (1 − t ) e  te −t −t Characteristic equation: = y, −y= x (b) State variables: x 1 ∆ (s ) = ( s +1) = 0 2 x 2 = y dy + dt State equations: dx dt 1 = dy dt = x 2 dx − x1 2 2 dt d = 2 y dt dy + 2 dt = −y − dy +r = −x2 +r dt  dx1   dt   −1 2   x1  0   =   + r  dx2   0 −1  x2  1   dt  State transition matrix:  1 s + 1 −2   s + 1 Φ (s ) =  =  0 s + 1   0  ( s + 1)  1   s +1  −2 −1 2 (c) Characteristic equation: ∆ (s ) = ( s +1) = 0 φ (t ) = e −t 0  − te e −t −t    2 5-20 (a) State transition matrix: ω  s − σ sI − A =  −ω s − σ  ( sI − A ) −1 which is the same as in part (a).

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Automatic Control Systems Solutions Manual Kuo Golnaragh by Benjamin Kuo


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